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Sagot :
Given the function:
[tex]f(x)=3x^5-10x^3[/tex]1. You need to find the first derivative.
Remember the Power Rule Derivative:
[tex]\frac{d}{dx}(x^n)=nx^{n-1}[/tex]Then:
[tex]f^{\prime}(x)=(3)(5)x^{5-1}-(3)(10)x^{3-1}[/tex][tex]f^{\prime}(x)=15x^4-30x^2[/tex]2. Make the first derivative equal to zero:
[tex]15x^4-30x^2=0[/tex]3. Solve for "x":
- Identify the Greatest Common Factor (the largest factor the terms have in common):
[tex]GCF=15x^2[/tex]- Factor the Greatest Common Factor out:
[tex]15x^2(x^2-2)=0[/tex]- Notice that you can divide the equation into two parts and solve for "x":
[tex]\begin{gathered} 15x^2=0\Rightarrow x=\frac{15}{0}\Rightarrow x=0 \\ \\ x^2-2=0\Rightarrow x=\pm\sqrt{2}\Rightarrow\begin{cases}x={-\sqrt{2}} \\ x={\sqrt{2}}\end{cases} \end{gathered}[/tex]4. Find the second derivate by derivating the first derivative:
[tex]f^{^{\prime}\prime}(x)=(15)(4)x^{4-1}-(30)(2)x^{2-1}[/tex][tex]f^{^{\prime}\prime}(x)=60x^3-60x^[/tex]5. Substitute the values of "x" found in Step 3 into the second derivative and evaluate:
[tex]f^{^{\prime}\prime}(0)=60(0)^3-60(0)=0[/tex][tex]f^{^{\prime}\prime}(-\sqrt{2})=60(-\sqrt{2})^3-60(-\sqrt{2})\approx-84.9[/tex][tex]f^{^{\prime}\prime}(\sqrt{2})=60(\sqrt{2})^3-60(\sqrt{2})\approx84.9[/tex]6. According to the Second Derivative Test:
- If:
[tex]f^{\prime}^{\prime}(x)>0[/tex]Then the function has a local minimum at that x-value.
- If:
[tex]f^{\prime\prime}(x)<0[/tex]Then the function has a local maximum at that x-value.
In this case:
[tex]f(-\sqrt{2})<0[/tex][tex]f(\sqrt{2})>0[/tex]Hence, the answer is:
- Local Minimum at:
[tex]x=\sqrt{2}[/tex]- Local Maximum at:
[tex]x=-\sqrt{2}[/tex]
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